Answer:
The ratio is
![0.67 : 1](https://img.qammunity.org/2021/formulas/chemistry/high-school/bv17943xzdhiyxhbajkqe66uvzsq4g4fyo.png)
Step-by-step explanation:
From the question we are told that
The activation energy separating the equatorial and the axial form is
![\Delta E = 0.24\ kcal/mol](https://img.qammunity.org/2021/formulas/chemistry/high-school/7rmg4wpf8jqiolmfb4eomxnyj6gghx76z4.png)
Generally according to the Boltzmann distribution , the relationship between concentration(in terms of number of molecule) of the equatorial form and the concentration of the axial form is mathematically represented as
![N = N_o e^{-(\Delta E )/(RT) }](https://img.qammunity.org/2021/formulas/chemistry/high-school/cyr6pom0gyy6glo9dwglrd29v70w9o3guc.png)
Here
is the number of molecule in equatorial form and N is the number of the molecules in the axial form.
R is the gas constant with value
![R = 1.987 *10^(-3) \ kcal\cdot mol^(-1) \cdot k^(-1)](https://img.qammunity.org/2021/formulas/chemistry/high-school/2l35qytqiy8qj4eqpbydwvo531c1bp6ln8.png)
T is the temperature of the room with value
![T = 25^oC = 298 \ K](https://img.qammunity.org/2021/formulas/chemistry/high-school/jen7hyhfqtxyvuiafgasp8r61ahdj92xom.png)
So
![(N)/(N_o) = e^{-(0.24 )/(1.987*10^(-3) * 298) }](https://img.qammunity.org/2021/formulas/chemistry/high-school/akfucvo3tehthnq9gaye52xssaags9vc8f.png)
=>
![(N)/(N_o) = 0.67](https://img.qammunity.org/2021/formulas/chemistry/high-school/pbrq7je304nl46oapgunta1gaasonhyrch.png)
So the ratio of the concentration of equatorial form to the axial form is
![0.67 : 1](https://img.qammunity.org/2021/formulas/chemistry/high-school/bv17943xzdhiyxhbajkqe66uvzsq4g4fyo.png)